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Multiple modulus inequalities

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Multiple modulus inequalities  [#permalink]

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New post 19 Dec 2009, 01:01
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Hi guys

Need to know how to solve multiple modulus inequalities

Ex - Solve |x-10| - |2x+3| <= |5x-10|

I do understand that we need to first find the roots, which happen to be 10, -3/2 and 2 in this case

we can then divide them into intervals -

x<-3/2
-3/2<x<2
2<x<10
x>10

what do we do post this division into intervals?
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Re: Multiple modulus inequalities  [#permalink]

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New post 19 Dec 2009, 02:47
3
zaarathelab wrote:
Hi guys

Need to know how to solve multiple modulus inequalities

Ex - Solve |x-10| - |2x+3| <= |5x-10|

I do understand that we need to first find the roots, which happen to be 10, -3/2 and 2 in this case

we can then divide them into intervals -

x<-3/2
-3/2<x<2
2<x<10
x>10

what do we do post this division into intervals?


If we dot he way you are proposing:

Yes, there are 3 check point at which absolute values in the example change their sign: -3/2, 2, and 10 (te values of x for which the expression in || equals to zero). Then you should expand the modulus in these ranges:

A. \(x<-\frac{3}{2}\) --> \(|x-10|-|2x+3|\leq{|5x-10|}\) --> \(-(x-10)+(2x+3)\leq{-(5x-10)}\) --> \(6x\leq{-3}\) --> \(x\leq{-\frac{1}{2}\) --> as we considering the range when \(x<-\frac{3}{2}\), then finally we should write \(x<-\frac{3}{2}\):
----(\(-\frac{3}{2}\))---------------------------------

B. \(-\frac{3}{2}\leq{x}\leq{2}\) --> \(-(x-10)-(2x+3)\leq{-(5x10)}\) --> \(2x<=3\) --> \(x<={\frac{3}{2}\) --> \({-\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\);
----(\(-\frac{3}{2}\))----(\(\frac{3}{2}\))-------------------------

C. \(2<x<10\) --> \(-(x-10)-(2x+3)\leq{5x-10}\) --> \(17\leq{8x}\) --> \(x\geq{\frac{17}{8}}\) --> \({\frac{17}{8}}\leq{x}<10\);
---------------------(\(\frac{17}{8}\))--------(\(10\))----

D. \(x\geq{10}\) --> \((x-10)-(2x+3)\leq{5x-10}\) --> \(-3\leq{6x\) --> \(x\geq{-\frac{1}{2}}\) --> \(x\geq{10}\);
---------------------(\(\frac{17}{8}\))--------(\(10\))----

After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are:
\(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\)
----(\(-\frac{3}{2}\))----(\(\frac{3}{2}\))----(\(\frac{17}{8}\))--------(\(10\))-----


One important detail when you define the critical points (10, -3/2 and 2) you should include them in either of intervals you check. So when you wrote:
x<-3/2
-3/2<x<2
2<x<10
x>10

You should put equal sign for each of the critical point in either interval. For example:
x=<-3/2
-3/2<x<=2
2<x<1=0
x>10

Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality.
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Re: Multiple modulus inequalities  [#permalink]

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New post 01 Feb 2012, 01:31
Could somebody explain this in detail, please?

I didn't get the concept of roots here, it doesn't make the equation zero - isn't it?
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Re: Multiple modulus inequalities  [#permalink]

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New post 01 Feb 2012, 01:47
bsaikrishna wrote:
Could somebody explain this in detail, please?

I didn't get the concept of roots here, it doesn't make the equation zero - isn't it?


When x<=0 then |x|=x and if x>0 then |x|=-x.

For |x-10|: when x<=10 then |x-10|=-(x-10) (because if x<=10 then x-10<=0) and when x>10 then |x-10|=x-10 (because if x>10 then x-10>0). So, zaarathelab calls "roots" (you can also call "critical point" or "check point") the values of x for which the expression in || equals to zero: in two different cases when x<that value and x>that value the absolute values expands with different sign (it switches the sign at this value).

Check Absolute Values chapter of Math Book for more: math-absolute-value-modulus-86462.html

Hope it helps.
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Re: Multiple modulus inequalities  [#permalink]

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New post Updated on: 01 Feb 2012, 02:05
Hi,

I solved the modulus by taking the positive & negative scenarios of each modulus. I solved them separately & arrived at 1/2 <= x & x<= -1/2.

Am I on the right track or messed up some where??
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Originally posted by boomtangboy on 01 Feb 2012, 01:55.
Last edited by boomtangboy on 01 Feb 2012, 02:05, edited 1 time in total.
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Re: Multiple modulus inequalities  [#permalink]

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New post 01 Feb 2012, 01:59
boomtangboy wrote:
Hi,

I solved the modulus by taking the positive & negative scenarios of each modulus. I solved them separately & arrived at 1/2 <= x & x<= -1/2.

Am I on the right track or messed up some where??


The answer to the initial question is x<=3/2 or 17/8<=x, so "1/2 <= x & x<= -1/2" is not right. Check my previous post: you should check different cases for different ranges to get the correct answer.
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Re: Multiple modulus inequalities  [#permalink]

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New post 04 Oct 2012, 10:45
Bunuel wrote:
zaarathelab wrote:
Hi guys

Need to know how to solve multiple modulus inequalities

Ex - Solve |x-10| - |2x+3| <= |5x-10|

I do understand that we need to first find the roots, which happen to be 10, -3/2 and 2 in this case

we can then divide them into intervals -

x<-3/2
-3/2<x<2
2<x<10
x>10

what do we do post this division into intervals?


If we dot he way you are proposing:

Yes, there are 3 check point at which absolute values in the example change their sign: -3/2, 2, and 10 (te values of x for which the expression in || equals to zero). Then you should expand the modulus in these ranges:

A. \(x<-\frac{3}{2}\) --> \(|x-10|-|2x+3|\leq{|5x-10|}\) --> \(-(x-10)+(2x+3)\leq{-(5x-10)}\) --> \(6x\leq{-3}\) --> \(x\leq{-\frac{1}{2}\) --> as we considering the range when \(x<-\frac{3}{2}\), then finally we should write \(x<-\frac{3}{2}\):
----(\(-\frac{3}{2}\))---------------------------------

B. \(-\frac{3}{2}\leq{x}\leq{2}\) --> \(-(x-10)-(2x+3)\leq{-(5x10)}\) --> \(2x<=3\) --> \(x<={\frac{3}{2}\) --> \({-\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\);
----(\(-\frac{3}{2}\))----(\(\frac{3}{2}\))-------------------------

C. \(2<x<10\) --> \(-(x-10)-(2x+3)\leq{5x-10}\) --> \(17\leq{8x}\) --> \(x\geq{\frac{17}{8}}\) --> \({\frac{17}{8}}\leq{x}<10\);
---------------------(\(\frac{17}{8}\))--------(\(10\))----

D. \(x\geq{10}\) --> \((x-10)-(2x+3)\leq{5x-10}\) --> \(-3\leq{6x\) --> \(x\geq{-\frac{1}{2}}\) --> \(x\geq{10}\);
---------------------(\(\frac{17}{8}\))--------(\(10\))----

After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are:
\(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\)
----(\(-\frac{3}{2}\))----(\(\frac{3}{2}\))----(\(\frac{17}{8}\))--------(\(10\))-----


One important detail when you define the critical points (10, -3/2 and 2) you should include them in either of intervals you check. So when you wrote:
x<-3/2
-3/2<x<2
2<x<10
x>10

You should put equal sign for each of the critical point in either interval. For example:
x=<-3/2
-3/2<x<=2
2<x<1=0
x>10

Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality.



Hi Brunel,

How do we decide the signs in front of the modulus in the equation. :(
Eg:: |x-10| - |2x+3| <= |5x-10|......say we are looking in an interval - 3/2 to 2. then based on what concentration do we put +or - sign in front of the modulus.


and please explain >> step 1----------x = sqrt (x^2) ?
< step w--------------- => x = |x| ? How do this. please explain concept behind this math.

Thanks!!
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Re: Multiple modulus inequalities  [#permalink]

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New post 05 Oct 2012, 03:50
1
Maverick04308 wrote:
Hi Brunel,

How do we decide the signs in front of the modulus in the equation. :(
Eg:: |x-10| - |2x+3| <= |5x-10|......say we are looking in an interval - 3/2 to 2. then based on what concentration do we put +or - sign in front of the modulus.


and please explain >> step 1----------x = sqrt (x^2) ?
< step w--------------- => x = |x| ? How do this. please explain concept behind this math.

Thanks!!


Explained here: multiple-modulus-inequalities-88174.html#p1037473

Hope it helps.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Multiple modulus inequalities  [#permalink]

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New post 11 Jan 2019, 08:49
Bunuel wrote:
zaarathelab wrote:
Hi guys

Need to know how to solve multiple modulus inequalities

Ex - Solve |x-10| - |2x+3| <= |5x-10|

I do understand that we need to first find the roots, which happen to be 10, -3/2 and 2 in this case

we can then divide them into intervals -

x<-3/2
-3/2<x<2
2<x<10
x>10


Bun

what do we do post this division into intervals?


If we dot he way you are proposing:

Yes, there are 3 check point at which absolute values in the example change their sign: -3/2, 2, and 10 (te values of x for which the expression in || equals to zero). Then you should expand the modulus in these ranges:

A. \(x<-\frac{3}{2}\) --> \(|x-10|-|2x+3|\leq{|5x-10|}\) --> \(-(x-10)+(2x+3)\leq{-(5x-10)}\) --> \(6x\leq{-3}\) --> \(x\leq{-\frac{1}{2}\) --> as we considering the range when \(x<-\frac{3}{2}\), then finally we should write \(x<-\frac{3}{2}\):
----(\(-\frac{3}{2}\))---------------------------------

B. \(-\frac{3}{2}\leq{x}\leq{2}\) --> \(-(x-10)-(2x+3)\leq{-(5x10)}\) --> \(2x<=3\) --> \(x<={\frac{3}{2}\) --> \({-\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\);
----(\(-\frac{3}{2}\))----(\(\frac{3}{2}\))-------------------------

C. \(2<x<10\) --> \(-(x-10)-(2x+3)\leq{5x-10}\) --> \(17\leq{8x}\) --> \(x\geq{\frac{17}{8}}\) --> \({\frac{17}{8}}\leq{x}<10\);
---------------------(\(\frac{17}{8}\))--------(\(10\))----

D. \(x\geq{10}\) --> \((x-10)-(2x+3)\leq{5x-10}\) --> \(-3\leq{6x\) --> \(x\geq{-\frac{1}{2}}\) --> \(x\geq{10}\);
---------------------(\(\frac{17}{8}\))--------(\(10\))----

After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are:
\(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\)
----(\(-\frac{3}{2}\))----(\(\frac{3}{2}\))----(\(\frac{17}{8}\))--------(\(10\))-----


One important detail when you define the critical points (10, -3/2 and 2) you should include them in either of intervals you check. So when you wrote:
x<-3/2
-3/2<x<2
2<x<10
x>10

You should put equal sign for each of the critical point in either interval. For example:
x=<-3/2
-3/2<x<=2
2<x<1=0
x>10

Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality.


Bunuel,

I believe that I follow along entirely until you reach the point where you are determining the viability of the solutions.

How have you checked the viability of the solutions? In this tutorial:https://gmatclub.com/forum/math-absolute-value-modulus-86462.html

It seems to teach that one must simply determine if the solution is within the range they are checking and if not, then it is not viable.

So, in this example, would x<=-1/2 satisfy the criteria of x<-3/2 because x would have to be less than -1/2 to be less than 3/2?

Also, how do you conclude that the inequality does not hold between 3/2 and 17/8? Again, I was able to arrive at the same numbers, but I am uncertain how you figured that the inequality does not hold in that interval.

Sorry as I am a bit lost! Thank you in advance for your help!


Best,


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Re: Multiple modulus inequalities  [#permalink]

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New post 12 Jan 2019, 02:55
1
nch2024 wrote:
Bunuel wrote:
zaarathelab wrote:
Hi guys

Need to know how to solve multiple modulus inequalities

Ex - Solve |x-10| - |2x+3| <= |5x-10|

I do understand that we need to first find the roots, which happen to be 10, -3/2 and 2 in this case

we can then divide them into intervals -

x<-3/2
-3/2<x<2
2<x<10
x>10


Bun

what do we do post this division into intervals?


If we dot he way you are proposing:

Yes, there are 3 check point at which absolute values in the example change their sign: -3/2, 2, and 10 (te values of x for which the expression in || equals to zero). Then you should expand the modulus in these ranges:

A. \(x<-\frac{3}{2}\) --> \(|x-10|-|2x+3|\leq{|5x-10|}\) --> \(-(x-10)+(2x+3)\leq{-(5x-10)}\) --> \(6x\leq{-3}\) --> \(x\leq{-\frac{1}{2}\) --> as we considering the range when \(x<-\frac{3}{2}\), then finally we should write \(x<-\frac{3}{2}\):
----(\(-\frac{3}{2}\))---------------------------------

B. \(-\frac{3}{2}\leq{x}\leq{2}\) --> \(-(x-10)-(2x+3)\leq{-(5x10)}\) --> \(2x<=3\) --> \(x<={\frac{3}{2}\) --> \({-\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\);
----(\(-\frac{3}{2}\))----(\(\frac{3}{2}\))-------------------------

C. \(2<x<10\) --> \(-(x-10)-(2x+3)\leq{5x-10}\) --> \(17\leq{8x}\) --> \(x\geq{\frac{17}{8}}\) --> \({\frac{17}{8}}\leq{x}<10\);
---------------------(\(\frac{17}{8}\))--------(\(10\))----

D. \(x\geq{10}\) --> \((x-10)-(2x+3)\leq{5x-10}\) --> \(-3\leq{6x\) --> \(x\geq{-\frac{1}{2}}\) --> \(x\geq{10}\);
---------------------(\(\frac{17}{8}\))--------(\(10\))----

After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are:
\(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\)
----(\(-\frac{3}{2}\))----(\(\frac{3}{2}\))----(\(\frac{17}{8}\))--------(\(10\))-----


One important detail when you define the critical points (10, -3/2 and 2) you should include them in either of intervals you check. So when you wrote:
x<-3/2
-3/2<x<2
2<x<10
x>10

You should put equal sign for each of the critical point in either interval. For example:
x=<-3/2
-3/2<x<=2
2<x<1=0
x>10

Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality.


Bunuel,

I believe that I follow along entirely until you reach the point where you are determining the viability of the solutions.

How have you checked the viability of the solutions? In this tutorial:https://gmatclub.com/forum/math-absolute-value-modulus-86462.html

It seems to teach that one must simply determine if the solution is within the range they are checking and if not, then it is not viable.

So, in this example, would x<=-1/2 satisfy the criteria of x<-3/2 because x would have to be less than -1/2 to be less than 3/2?

Also, how do you conclude that the inequality does not hold between 3/2 and 17/8? Again, I was able to arrive at the same numbers, but I am uncertain how you figured that the inequality does not hold in that interval.

Sorry as I am a bit lost! Thank you in advance for your help!


Best,


NCH2024


-3/2 < -1/2, so x's which are less than -3/2 are also less than -1/2.

----(\(-\frac{3}{2}\))------------\((-\frac{1}{2})\)---------------------
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Multiple modulus inequalities  [#permalink]

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New post 15 Jan 2019, 05:34
Bunuel wrote:
Bunuel wrote:
Hi guys

Need to know how to solve multiple modulus inequalities

Ex - Solve |x-10| - |2x+3| <= |5x-10|

I do understand that we need to first find the roots, which happen to be 10, -3/2 and 2 in this case

we can then divide them into intervals -

x<-3/2
-3/2<x<2
2<x<10
x>10


-3/2 < -1/2, so x's which are less than -3/2 are also less than -1/2.

----(\(-\frac{3}{2}\))------------\((-\frac{1}{2})\)---------------------
[/quote]

Thank you, Bunuel!
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Re: Multiple modulus inequalities &nbs [#permalink] 15 Jan 2019, 05:34
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